# Integral of sqrt(1 + e^(2x))

• Feb 8th 2011, 12:43 PM
jsel21
Integral of sqrt(1 + e^(2x))
Integral of sqrt(1 + e^(2x))

u = 1 + e^(2x) ??

Please point me in the right direction. Thank you.
• Feb 8th 2011, 12:46 PM
dwsmith
Quote:

Originally Posted by jsel21
Integral of sqrt(1 + e^(2x))

u = 1 + e^(2x) ??

Please point me in the right direction. Thank you.

$1+\tan^2=\sec^2$

$e^{2x}=\left(e^x\right)^2$

$e^x=\tan\theta$

Try this.
• Feb 8th 2011, 02:58 PM
mr fantastic
Quote:

Originally Posted by jsel21
Integral of sqrt(1 + e^(2x))

u = 1 + e^(2x) ??

Please point me in the right direction. Thank you.

integrate Sqrt&#91;1 &#43; Exp&#91;2x&#93;&#93; - Wolfram|Alpha+

Click on Show steps.
• Feb 8th 2011, 06:06 PM
TheCoffeeMachine
\displaystyle \begin{aligned} \int \sqrt{1+e^{2x}}\;{dx} & = \int\frac{\sqrt{1+e^{2x}}\sqrt{1+e^{2x}}}{\sqrt{1+ e^{2x}}}\;{dx} \\&= \int\frac{1+e^{2x}}{\sqrt{1+e^{2x}}}\;{dx} \\& = \int\frac{1}{\sqrt{1+e^{2x}}}\;{dx}+\int\frac{e^{2 x}}{\sqrt{1+e^{2x}}}\;{dx}
\\& = I_{1}+\int \frac{(1+e^{2x})'}{2\sqrt{1+e^{2x}}}\;{dx} \\& = I_{1}+\sqrt{1+e^{2x}}+k.\end{aligned}

Where

$\displaystyle I_{1} = \int\frac{1}{\sqrt{1+e^{2x}}}\;{dx}.$

Let $t = \sqrt{1+e^{2x}}$, so that $dx = \frac{\sqrt{1+e^{2x}}}{e^{2x}}\;{dt} = \frac{t}{t^2-1}\;{dt}$, then:

\displaystyle \begin{aligned} I_{1} & = \int\frac{t}{t(t^2-1)}\;{dt} \\& = \int\frac{1}{(t-1)(t+1)}\;{dt} \\ & = \frac{1}{2}\int \frac{(t+1)-(t-1)}{(t+1)(t-1)}\;{dt} \\& = \frac{1}{2}\int\frac{1}{t-1}\;{dt}-\frac{1}{2}\int\frac{1}{t+1}\;{dt} \\& = \frac{1}{2}\ln(t-1)-\frac{1}{2}\ln(t+1)+k \\ & = \frac{1}{2}\ln\left\{\frac{t-1}{t+1}\right\}+k \\ & = \frac{1}{2}\ln\left\{\frac{\sqrt{1+e^{2x}}-1}{\sqrt{1+e^{2x}}+1}\right\}+k_{1}. \end{aligned}

Thus $\displaystyle \boxed{\int \sqrt{1+e^{2x}}\;{dx} = \frac{1}{2}\ln\left\{\frac{\sqrt{1+e^{2x}}-1}{\sqrt{1+e^{2x}}+1}\right\}+\sqrt{1+e^{2x}}+k}.$
• Feb 8th 2011, 07:10 PM
Prove It
Quote:

Originally Posted by dwsmith
$1+\tan^2=\sec^2$

$e^{2x}=\left(e^x\right)^2$

$e^x=\tan\theta$

Try this.

Interesting...

$\displaystyle \int{\sqrt{1 + e^{2x}}\,dx}$

Make the substitution $\displaystyle e^x = \tan{\theta} \implies x = \ln{(\tan{\theta})} \implies dx = \frac{\sec^2{\theta}}{\tan{\theta}}\,d\theta$ and the integral becomes

$\displaystyle \int{\sqrt{1 + \tan^2{\theta}}\,\frac{\sec^2{\theta}}{\tan{\theta }}\,d\theta}$

$\displaystyle = \int{\sec{\theta}\,\frac{\sec^2{\theta}}{\tan{\the ta}}\,d\theta}$

$\displaystyle = \int{\sec^3{\theta}\cot{\theta}\,d\theta}$

$\displaystyle = \int{\frac{1}{\cos^3{\theta}}\,\frac{\cos{\theta}} {\sin{\theta}}\,d\theta}$

$\displaystyle = \int{\frac{1}{\sin{\theta}\cos^2{\theta}}\,d\theta }$

$\displaystyle = \int{\frac{\sin{\theta}}{\sin^2{\theta}\cos^2{\the ta}}\,d\theta}$

$\displaystyle = \int{\frac{\sin{\theta}}{(1 - \cos^2{\theta})\cos{\theta}}\,d\theta}$

$\displaystyle = \int{\frac{-\sin{\theta}}{(\cos^2{\theta} - 1)\cos{\theta}}\,d\theta}$

$\displaystyle = \int{\frac{-\sin{\theta}}{(\cos{\theta} - 1)(\cos{\theta} + 1)\cos{\theta}}\,d\theta}$.

And now making the substitution $\displaystyle u = \cos{\theta} \implies du = -\sin{\theta}\,d\theta$ the integral becomes

$\displaystyle \int{\frac{1}{(u-1)(u+1)u}\,du}$

which can be solved using partial fractions :)
• Feb 8th 2011, 07:12 PM
dwsmith
Quote:

Originally Posted by Prove It
Interesting...

$\displaystyle \int{\sqrt{1 + e^{2x}}\,dx}$

Make the substitution $\displaystyle e^x = \tan{\theta} \implies x = \ln{(\tan{\theta})} \implies dx = \frac{\sec^2{\theta}}{\tan{\theta}}\,d\theta$ and the integral becomes

$\displaystyle \int{\sqrt{1 + \tan^2{\theta}}\,\frac{\sec^2{\theta}}{\tan{\theta }}\,d\theta}$

$\displaystyle = \int{\sec{\theta}\,\frac{\sec^2{\theta}}{\tan{\the ta}}\,d\theta}$

$\displaystyle = \int{\sec^3{\theta}\cot{\theta}\,d\theta}$

$\displaystyle = \int{\frac{1}{\cos^3{\theta}}\,\frac{\cos{\theta}} {\sin{\theta}}\,d\theta}$

$\displaystyle = \int{\frac{1}{\sin{\theta}\cos^2{\theta}}\,d\theta }$

$\displaystyle = \int{\frac{\sin{\theta}}{\sin^2{\theta}\cos^2{\the ta}}\,d\theta}$

That was the first thing that came to mind.
• Feb 8th 2011, 07:48 PM
jsel21
Thanks guys, these posts were very helpful.
• Feb 8th 2011, 07:54 PM
Krizalid
$\displaystyle\int{\sqrt{1+{{e}^{2x}}}\,dx}=\int{\f rac{\sqrt{1+{{e}^{2x}}}{{e}^{2x}}}{{{e}^{2x}}}\,dx }=\int{\frac{{{t}^{2}}}{{{t}^{2}}-1}\,dt},$ there's no much here, just clearing out the clutter and it's easy.
• Feb 8th 2011, 10:12 PM
Abu-Khalil
Also

$\displaystyle\int\frac{dt}{\sin t\cos^2 t}=\int\frac{\sin^2 t+\cos^2}{\sin t\cos^2 t}dt=\int\frac{\sin t}{\cos^2 t}dt+\int\csc tdt$
• Feb 9th 2011, 02:03 AM
ice_syncer
take u^2 = 1 + e^(2x)
so 2udu = 2e^(2x)
which is 2(u^2-1)dx
2udu = 2(u^2-1)dx
dx = u/(u^2-1) du
put it in the question.. it will be u^2/(u^2-1) du = (u^2 - 1 + 1)/(u^2-1)du
• Feb 9th 2011, 09:19 AM
Krizalid
you did the same as me.
• Feb 9th 2011, 10:57 AM
ice_syncer
hmm.. sorry
• Feb 9th 2011, 01:57 PM
topsquark
Quote:

Originally Posted by Prove It
$\displaystyle \int{\sqrt{1 + \tan^2{\theta}}\,\frac{\sec^2{\theta}}{\tan{\theta }}\,d\theta}$

$\displaystyle = \int{\sec{\theta}\,\frac{\sec^2{\theta}}{\tan{\the ta}}\,d\theta}$

I hate it when trig substitutions do this kind of thing.
$sec( \theta ) = \pm \sqrt{1 + tan^2 ( \theta )}$

But the original integrand is positive definite, so we can discard the negative solution. Correct? Or is it more complicated than that?

-Dan
• Feb 9th 2011, 03:43 PM
Krizalid
In this case, on Prove's substitution, that forces to $0\le\theta<\dfrac\pi2$ and we have that $|\sec\theta|=\sec\theta.$

Of course we coulda taken $\pi\le\theta<\dfrac32\pi$ but that makes $\sec\theta$ non positive.

It all depends on how you fix the angle, but that's more a matter of dealing with definite integrals.